Proximity algorithms for nearly-doubling spaces

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Proximity algorithms for nearly-doubling spaces

• Lee-Ad Gottlieb
• Robert Krauthgamer
• Weizmann Institute

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Proximity problems

• In arbitrary metric space, some proximity
  problems are hard
• For example, the nearest neighbor search problem
  requires T(n) time
• The doubling dimension
• parameterizes the bad
• case

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Doubling Dimension

• Definition Ball B(x,r) all points within
  distance r from x.
• The doubling constant (of a metric M) is the
  minimum value gt0 such that every ball can be
  covered by balls of half the radius
• First used by Ass-83, algorithmically by
  Cla-97.
• The doubling dimension is dim(M)log (M)
  GKL-03
• A metric is doubling if its doubling dimension is
  constant
• Packing property of doubling spaces
• A set with diameter D and min. inter-point
• distance a, contains at most
• (D/a)O(log) points

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Applications

• In the past few years, many algorithmic tasks
  have been analyzed via the doubling dimension
• For example, approximate nearest neighbor search
  can be executed in time O(1) log n
• Some other algorithms analyzed via the doubling
  dimension
• Nearest neighbor search KL-04, BKL-06, CG-06
• Clustering Tal-04, ABS-08, FM-10
• Spanner construction GGN-06, CG-06, DPP-06,
  GR-08
• Routing KSW-04, Sil-05, AGGM-06, KRXY-07,
  KRX-08
• Travelling Salesperson Tal-04
• Machine learning BLL-09, GKK-10
• Message This is an active line of research

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Problem

• Most algorithms developed for doubling spaces are
  not robust
• Algorithmic guarantees dont hold for
  nearly-doubling spaces
• If a small fraction of the working set possesses
  high doubling dimension, algorithmic performance
  degrades.
• This problem motivates the following key task
• Given an n-point set S and target dimension d
• Remove from S the fewest number of points so that
  the remaining set has doubling dimension at most
  d

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Two paradigms

• How can removing a few bad points help? Two
  models
• 1. Ignore the bad points
• Outlier detection.
• GHPT-05 cluster based on similarity, seek a
  large subset with low intrinsic dimension.
• Algorithms with slack. Throw bad points into the
  slack
• KRXY-07 gave a routing algorithm with
  guarantees for most of the input points.
• FM-10 gave a kinetic clustering algorithm for
  most of the input points.
• GKK-10 gave a machine learning algorithm
  small subset doesnt interfere with learning

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Two paradigms

• How can removing a few bad points help? Two
  models
• 2. Tailor a different algorithm for the bad
  points
• Example Spanner construction. A spanner is an
  edge subset of the full graph
• Good points Low doubling dimension sparse
  spanner with nice properties (low stretch and
  degree)
• Bad points Take the full graph
• If the number of bad points is O(n.5), we have a
  spanner with O(n) edges

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Results

• Recall our key problem
• Given an n-point set S and target dimension d
• Remove from S the fewest number of points so that
  the remaining set has doubling dimension at most
  d
• This problem is NP-hard
• Even determining the doubling dimension of a
  point set exactly is NP-hard!
• Proof on the next slide
• But the doubling dimension can be approximated
  within a constant factor
• Our contribution bicriteria approximation
  algorithm
• In time 2O(d) n3, we remove a number of points
  arbitrarily close to optimal, while achieving
  doubling dimension 4d O(1)
• We can also achieve near-linear runtime, at the
  cost of slightly higher dimension

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Warm up

• Lemma It is NP-hard to determine the doubling
  dimension of a set S
• Reduction from vertex cover with bounded degree
  ? n½.
• the size of any vertex cover is at least n½.
• Construction A set S of n points corresponding
  to the vertex set V.
• Let d(u,v) ½ if the cor. vertices are
  connected by an edge
• Let d(u,v) 1 if the cor. vertices arent
  connected
• Analysis
• Any subset of S found in a ball of radius ½ has
  at most n½ points - degree of original graph
• S is a ball of radius 1. The minimum covering of
  all of S with balls of radius ½ is equal to the
  minimum vertex cover of V.
• Note reduction preserves hardness of
  approximation
• Corollary It is NP-hard to determine if removing
  k points from S can leave a set with doubling
  dimension d.
• So our problem is hard as well.

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Bicriteria algorithm

• Recall that he doubling constant (of a metric M)
  is
• the minimum value gt0 such that every r-radius
  ball can be covered by balls of half the radius
• Define the related notion of density constant as
• the minimum value mgt0 such that every r-radius
  ball contains at most m points at mutual
  interpoint distance r/2
• Nice property The density constant can only
  decrease under the removal of points, unlike the
  doubling constant.
• We can show that
• vm(S) (S) m(S)
• its NP-hard to compute the density constant
• (ratio-preserving reduction from independent set)

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Bicriteria algorithm

• We will give a bicriteria algorithm for the
  density constant. Problem statement
• Given an n-point set S and target density
  constant m
• Remove from S the fewest number of points so that
  the remaining set has density constant at most m
• A bicriteria algorithm for the density constant
  is itself a bicriteria algorithm for the doubling
  constant
• within a quadratic factor

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Witness set

• Given a set S, a subset S is a witness set for
  the density constant if
• All points are at interpoint distance at least
  r/2
• Note that S is a concise proof that the density
  constant of S is at least S
• Theorem Fix a value mlt m(S). A witness set of S
  of size at least vm can be found in time 2O(m)
  n3
• Proof outline
• For each point p and radius r define the r-ball
  of p.
• Greedily cover all points in the r-ball with
  disjoint balls of radius r/2.
• Then cover all points in each r/2 ball with
  disjoint balls of radius r/4.
• Since there exists in S a witness set of size
  m(S), there exists a p and r so that
• either there are vm(S) r/2 balls, and these form
  a witness set, or
• one r/2 ball covers vm(S) r/4 balls, and these
  form a witness set.

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Bicriteria algorithm

• Recall our problem
• Given an n-point set S and target density
  constant m
• Remove from S the fewest number of points so that
  the remaining set has density constant at most m
• Our bricriteria solution
• Let k be the true answer (the minimum number of
  points that must be removed).
• We remove k c/(c-1) points and the remaining set
  has density constant c2m2

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Bicriteria algorithm

• Algorithm
• Run the subroutine to identify a witness set of
  size at least cm
• Remove it
• Repeat
• Analysis
• The density constant of the resulting set is not
  greater than c2m2
• since we terminated without finding a witness set
  of size at least cm
• Every time a witness set of size wgtcm is
  removed by our algorithm, the optimal algorithm
  must remove at least w-m points
• or else the true solution would have density
  constant greater than m
• It follows that are algorithm removes k w/(w-m)
  lt kc/(c-1) points

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Conclusion

• We conclude that there exists a bicriteria
  algorithm for the density constant
• We remove k c/(c-1) points and the remaining set
  has density constant c2m2
• It follows that there exists a bricriteria
  algorithm for the doubling constant
• We remove k c/(c-1) points and the remaining set
  has doubling constant c44